If $W_1=Y_1+2Y_2$ and $W_2=4Y_1-Y_2$ what is the joint distribution of $W_1$ and $W_2$ Let $Y_1$ and $Y_2$ be independent random variables with
$Y_1\sim N(1,3)$ and $Y_2 \sim N(2,5).$
If $W_1=Y_1+2Y_2$ and $W_2=4Y_1-Y_2$
what is the joint distribution of $W_1$ and $W_2$?
Is correct my procedure?
$E(W_1)=E(Y_1+2Y_2)=E(Y_1)+2E(Y_2)=1+2(2)=5$
$Var(W_1)=var(Y_1+2Y_2)=var(Y_1)+4var(Y_2)=3+4(5)=23$
$E(W_2)=E(4Y_1-Y_2)=4E(Y_1)-E(Y_2)=4(1)-2$
$Var(W_2)=var(4Y_1-Y_2)=16var(Y_1)-var(Y_2)=16(3)-5=43$
$f(w_1)=\frac{1}{\sqrt{2\pi}23}\epsilon^{\frac{-1}{2}(\frac{x-5}{23})^2}$
$f(w_2)=\frac{1}{\sqrt{2\pi}43}\epsilon^{\frac{-1}{2}(\frac{x-2}{43})^2}$
$f(w_1,w_2)=\frac{1}{\sqrt{2\pi}23}\epsilon^{\frac{-1}{2}(\frac{x-5}{23})^2} \cdot \frac{1}{\sqrt{2\pi}43}\epsilon^{\frac{-1}{2}(\frac{x-2}{43})^2} $
 A: "Mostly" right, good progress. But you aren't quite finished. 
Because $W_1$ and $W_2$ are both
influenced by the same $Y_1$ and $Y_2$, it seems intuitively
clear that you need to find the covariance also. Remember
that $Cov$ is linear in both its arguments and that the $Y_i$ are independent.
Here is a brief simulation with answers that should be
accurate to 2 (maybe 3) places. Answers are based on a million
performances of the experiment to sample $Y_1$ and $Y_2$ and
transform them to $W_1$ and $W_2.$ You can check the answers
you have (I see an error) and then the covariance that you need to find.
Finally, you need to look at the formula for the bivariate
normal density function.
 m = 10^6;  y1 = rnorm(m, 1, sqrt(3));  y2 = rnorm(m, 2, sqrt(5))
 w1 = y1 + 2*y2;  w2 = 4*y1 - y2
 mean(y1);  sd(y1);  mean(y2);  sd(y2);  cov(y1,y2)
 ## 0.9951616
 ## 1.732375
 ## 2.001808
 ## 2.235283
 ## -0.003585973  # Consistent with indep y1 and y2

 mean(w1);  sd(w1);  mean(w2);  sd(w2)
 ## 4.998779
 ## 4.792989
 ## 1.978838
 ## 7.283074
 sqrt(23)
 ## 4.795832
 sqrt(43)      
 ## 6.557439  # Better check formula for variance of difference

 cov(w1,y2); cor(w1, w2)
 ## 9.989395
 ## 0.05690479

Even if someone is eager to show off the right answers, you will
learn something by looking at these approximate answers, and
reviewing the necessary relationships. Please leave me a Comment
if there is something you can't figure out.
